When integer x^3 is divided by 256, the remainder is 0. Which if the

Question

When integer x^3 is divided by 256, the remainder is 0. Which if the following could be the remainder when x is divided by 256?

I. 2
II. 8
III. 20

A. None
B. I only
C. II only
D. III only
E. I, II, and III

Natashaairan · Answer

Please explain

X can have a possible value as 2 
As 2^3 will leave reminder 0

But 8 cube and 20 cube won’t

So x can only be 2 
2 when divided by 256

Will leave reminder 2

So I think the answer should be B ??

Posted from my mobile device

paridhi987 · Answer

But 8^3=512 which is divisible by 256?

Kinshook · Answer

paridhi987

When integer x^3 is divided by 256, the remainder is 0. Which if the following could be the remainder when x is divided by 256?

I. 2
II. 8
III. 20

x^3 = 256k = 2^8k = 2^6*4k

k is of the form 2m^3
x ^3 = 2^8*2m^3 = 2^9m^3
x = 2^3m = 8m

Case 1: x = 8m = 256k + 8; 256k = 8(m-1); When m = 1; x = 8; The remainder when x is divided by 256 = 8
Case 2: x = 8m = 256k + 2; 256k = 8m - 2 = 2(4m-1): Not feasible since 4m-1 is odd
Case 3: x = 8m = 256k + 20; 256k = 8m - 20 = 4(2m-5); Not feasible since 2m-5 is odd

A. None
B. I only
C. II only
D. III only
E. I, II, and III

IMO C

paridhi987 · Answer

Hi  Kinshook  , how do we get the step 'k is of the form 2m^3'?

adewale223 · Answer

Why is it not feasible if 4m-1 2m-5 are odd

stne · Answer

Given  x^3  =  256n

Hence  x  must be a multiple of  8

(I) 2 : For remainder to be  2

Check multiples of  8,16,24  ...etc not possible to get remainder of  2  when divided by  256

Reverse way : Any number such as  258  which gives a remainder of  2 , when divided by  256,  will not be divisible  256   when cubed.

\frac{(258)^3}{256}  = not an integer. This falsifies the question

(II) For remainder to be  8

If  x = 8  then  \frac{8^3}{256}  = integer, Also  \frac{8}{256}  = remainder  8 . True.

(III) : For remainder to be  20

Check multiples of  8,16,24  ...etc not possible to get remainder of  20  when divided by  256

Reverse way : Any number such as  276  which gives a remainder of  20,  when divided by  256 , will not be divisible by  256  when cubed.

\frac{(276)^3}{256}  = not an integer.This falsifies the question

Ans C

Hope it helped.

rish_dutton · Answer

x^3 = 256k = 2^8 * k where k is an integer

where 2^8 * k is a whole number. That means k MUST be of the form 2*m^3 where m is an integer

That means x^3 must be of the form:

x^3 
= 2^8 * (k)
= 2^8 * (2*m^3)
= 2^9 * m^3

That means x must be of the form 
2^3 * m or 8m

Now, for 8m, the values go like 0, 8, 16, 24 etc.

For any whole number a, the values of x can be also written as 256a + 8, 256a + 16, 256a + 24... each of which leaves a multiple of 8 as the remainder.

The only remainder that satisfies this condition in the three choices is 8.

Thus B.

Sudhanwa23 · Answer

Hi.. I'm sorry but can someone explain why the "k" in 256k must be of the form 2m^3 and why not just m^3?

Bunuel · Answer

x^3 = 2^8 * k

The left-hand side is the cube of an integer, so the right-hand side must also be the cube of an integer. This means the power of 2 in 2^8 must be completed to a multiple of 3, which can be done by multiplying it by another 2. So, k must be 2 multiplied by the cube of an integer:

x^3 = 2^8 * k = 2^8 * (2 * m^3) = 2^9 * m^3 = (2^3 * m)^3 = (integer)^3

Hope it helps.

Sudhanwa23 · Answer

Aw mann.. I gotta do better. Its clear now. Thank you so much!

Adit_ · Answer

Simplest way to solve is by noticing that you need 2^3 as the factor of x so that 2^3*3 = 2^9 would give remainder as 0 when divided by 256 or 2^8.
Any multiple of 8 would thus give a remainder of 0.
Thus 8y/256 would give remainder 0, 8, 16 and so on..
1 and 3 are thus out.
 2  is the only answer.

KarishmaB · Answer

x^3 is divisible by 256 = 2^8.
So x must be divisible by 2^3 = 8 at least. Hence we can make groups of 8 with nothing leftover from x.

When we divide x by 256, we will put together 32 groups of 8 to make a group of 256. Hence we may have no groups of 8 remaining, 1 group 8 remaining, 2 groups of 8 remaining etc till 31 groups of 8 remaining.
So the remainder will be 0 or 8 or 16 or 24 or 32 ... 248

Of the given options, only 8 is possible.

Answer (C)

Division and Remainders discussed here:

https://www.youtube.com/watch?v=A5abKfUBFSc

ankitray · Answer

It is given that  x^3  % 256 or  2^8  is zero.

So,  x^3 / 2^8  = k

=>  ((\frac{x}{2^2})^3)/4  = k

=> x can be 8, 16 , 24 , 32 ...... , 256, 264 , .....

=> Remainder when x is divided by 256 would be 8, 16, 24 , ......, 248, 0, 8, 16 , 24 ........

=> Distinct set of remainders will always be 0, 8, 16, 24 , ......... , 248.

And only option II matches this and Hence answer is Option C.

When integer x^3 is divided by 256, the remainder is 0. Which if the

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When integer x^3 is divided by 256, the remainder is 0. Which if the